Question: The distance between the two intersections of $x=y^4$ and $x+y^2=1$ is $\sqrt{u+v\sqrt5}$.  Find the ordered pair, $(u,v)$.
Solution: To find the $y$-coordinates of the intersections, substitute $y^4$ for $x$ in $x+y^2=1$ and solve for $y$, resulting in  \begin{align*}
y^4+y^2&=1 \\
\Rightarrow \qquad y^4+y^2-1&=0 \\
\Rightarrow \qquad y^2&=\frac{-1\pm\sqrt{1+4}}2=\frac{-1\pm\sqrt5}2\\
\end{align*}But $y^2$ is positive, so we reject $\frac{-1-\sqrt5}2$.  Therefore $y=\pm\sqrt{\frac{\sqrt5-1}2}$.


Using each of these coordinates to solve for $x$ gives us the intersections at  $\left(\frac{3-\sqrt5}2,\sqrt{\frac{\sqrt5-1}2}\right)$ and  $\left(\frac{3-\sqrt5}2,-\sqrt{\frac{\sqrt5-1}2}\right)$.  Using the distance formula, we have

\begin{align*}
&\sqrt{ \left(\frac{3-\sqrt5}2-\frac{3-\sqrt5}2\right)^2 + \left(\sqrt{\frac{\sqrt5-1}2}+\sqrt{\frac{\sqrt5-1}2}\right)^2 }\\
&\qquad=\sqrt{\left(2\sqrt{\frac{\sqrt5-1}2}\right)^2}\\
&\qquad=2\sqrt{\frac{\sqrt5-1}{2} }\\
&\qquad=\sqrt{2\sqrt5-2}.
\end{align*}So, $(u,v)=\boxed{(-2,2)}.$